A particle moves on a circle of radius 8 cm, centered at the origin, in the xy-plane (x and y measured in centimeters). It starts at the point (0,8) at time t=0 and moves counterclockwise, going once around the circle in 9 seconds.
What is the particle's speed?
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
Not the answer you are looking for? Search for more explanations.
A. Write a parameterization for the particle's motion
B. What is the particles speed?
Oh man, I should know this... I took that class last semester haha, I'll think about it.
There’s a lot going on in this problem.
We know that the particle has a period of 9.
Every 9 seconds, it returns to where it was 9 seconds ago.
And we probably suspect that the motion is parametrized by trig functions,
not only because they’re the only periodic functions we study a lot,
but also because the motion of the particle is circular, and circles relate to trig functions.
Let’s try to think of a periodic trig function whose period is 9.
Well, we know that sin t and cos t both are 2 pi –periodic.
Let’s focus on sin t. This is 2pi periodic.
Moreover, sin (2pi *t) is 1-periodic.
Its period is 1 second since we make the sine wave travel 2pi times faster.
Similarly, we see that sin ((2pi*t)/N)) is now N – periodic, for any integer N.
In particular, sin ((2pi*t)/9)) and cos ((2pi*t)/9)) are both trig functions with period 9.
Let x(t)and y(t) be the x and y position functions of the particle.
We see x(0) = 0 and y(0) = 8, since the particle starts at (0, 8).
You can figure out by guessing which trig function goes with x(t) and which one goes with y(t).
But here’s one way to approach it without trial and error.
Normally, if the particle had been placed on the x-axis at (8, 0),
we’d see that the particle is parametrized by
x(t) = 8 * cos ((2pi*t)/9)), y(t) = 8* sin ((2pi*t)/9)).
However, the position has been shifted and rotated 90 degrees.
Recall cos (x + pi/2) = cos x * cos pi/2 – sin x * sin pi/2 = - sin x
and sin (x + pi/2) = sin x * cos pi/2 + cos x * sin pi/2 = cos x.
So that means if we start at (8, 0) instead of (0, 8),
we must have rotated our graph by pi/2,
and so the position functions should actually be
x(t) = 8 * (-sin ((2pi*t)/9))) , y(t) = 8 * (cos ((2pi*t)/9))).
You can check to see that this works.
Now the speed is calculated kind of like the Pythagorean theorem.
Let x’(t) and y’(t) be the individual x and y velocity functions.
Then speed = sqrt( (x’)^2 + (y’)^2), just like if x’ and y’ were sides of a right triangle,
and the speed would be the hypotenuse.
If you calculate out x’ and y’, square them, add them,
notice that you have a sum of (cos x)^2 + (sin x)^2 which equals 1, and
take the square root, then you’ll get (2pi/9)*8 as your speed.
That kind of makes sense. The eight tells you it’s 8 times faster than a circle of radius 1.
The (2pi/9)~ 0.7 tells you it’s not at going as fast around the circle as it normally would for normal trig functions, ie. 2pi seconds.